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Math Help - Equation with primes in exponent

  1. #1
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    Equation with primes in exponent

    Let a,n be positive integers, and let p be a prime. Prove if
    (2^p) + (3^p) = (a^n), that n=1
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    Re: Equation with primes in exponent

    If p is a prime, then last digit of LHS=5 (p>2). So the last digit of a is 5, too...

    -------------------------------------------------------
    edit: p>2 added
    Last edited by godelproof; June 22nd 2011 at 07:53 PM.
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    Re: Equation with primes in exponent

    Quote Originally Posted by godelproof View Post
    If p is a prime, then last digit of LHS=5. So if n>1, then the last digit of a is 5, too...
    What do you mean by this? It's not true... unless you mean something other than what I think you mean.
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    Re: Equation with primes in exponent

    Quote Originally Posted by topspin1617 View Post
    What do you mean by this? It's not true... unless you mean something other than what I think you mean.


    Well, it is true if p is an odd prime and if by "last digit" he meant the units' digit:

    2^2=-1\!\!\pmod 5\Longrightarrow 2^{p=2n+1}=\left\{\begin{array}{rl}-2\!\!\!\pmod 5&\,\mbox{ if }n\mbox{ is odd}\\2\!\!\pmod 5&\,\mbox{ if }n\mbox{ is even}\end{array}\right. , and the same exactly

    happens with 3 instead of 2 , so in any case (with odd prime) the sum 2^p+3^p=0\!\!\pmod 5.

    Tonio
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    Re: Equation with primes in exponent

    Quote Originally Posted by tonio View Post
    Well, it is true if p is an odd prime and if by "last digit" he meant the units' digit:

    2^2=-1\!\!\pmod 5\Longrightarrow 2^{p=2n+1}=\left\{\begin{array}{rl}-2\!\!\!\pmod 5&\,\mbox{ if }n\mbox{ is odd}\\2\!\!\pmod 5&\,\mbox{ if }n\mbox{ is even}\end{array}\right. , and the same exactly

    happens with 3 instead of 2 , so in any case (with odd prime) the sum 2^p+3^p=0\!\!\pmod 5.

    Tonio
    Okay, this makes sense.

    I must be missing something obvious though.... I still don't see the solution here...
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    Re: Equation with primes in exponent

    Also notice that n must be an odd number...
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    Re: Equation with primes in exponent

    Quote Originally Posted by topspin1617 View Post
    Okay, this makes sense.

    I must be missing something obvious though.... I still don't see the solution here...
    This is NOT an easy question!
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    Re: Equation with primes in exponent

    Quote Originally Posted by topspin1617 View Post
    Okay, this makes sense.

    I must be missing something obvious though.... I still don't see the solution here...
    can we show that 5^2 \not | (2^p+3^p). if this is done then the question is solved.
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    Re: Equation with primes in exponent

    2^5+3^5=275
    275=5*5*11 \neq{a}^{n}, n>1
    i proved for any prime p>5, n=1

    Basically the proof runs like this: LHSmod(25)=0 iff the its last two digits \in{00,25,50,75}. Define B={00,25,50,75}. Define LT(x)=last two digits of x (so LT(100)=00, LT(234)=34, ect.). Consider f(k)=LT(2^k+3^k) for k=1,2,3,... Then we find f(k) \inB iff k=20m+5 or k=20m+15, for m=0,1,2,3,... Hence k cannot be a prime if LHSmod(25)=0. Hence k>5 cannot be a prime if n>2 on the RHS.

    -------------------------------------------------------------
    EDIT: RED LETTERS
    Last edited by godelproof; June 22nd 2011 at 11:32 PM.
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    Re: Equation with primes in exponent

    Now for the claim f(k) \inB iff k=20m+5 or k=20m+15, for m=0,1,2,3,... made above. See the attachment. All should be clear
    ------------------------------------------------------
    Edit: the smallest loop for LT is of length 20.
    Attached Thumbnails Attached Thumbnails Equation with primes in exponent-.jpg  
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    Re: Equation with primes in exponent

    EDIT:

    In the proof, (2^p) + (3^p) = (a^n) \Rightarrow n=1 is true if {p}={odd numbers}-{20m+5}-{20m+15} because for these p LHSmod(25) \neq0. So the set of p's for which n=1 is true is strictly larger than the prime set.

    Actually, I suspect that n=1 is true for ANY positive integer p! However a proof for that might be much more involved and I shall not attempt to prove it
    Last edited by godelproof; June 23rd 2011 at 03:31 AM.
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    Re: Equation with primes in exponent

    Quote Originally Posted by godelproof View Post
    EDIT:

    In the proof, (2^p) + (3^p) = (a^n) \Rightarrow n=1 is true if {p}={odd numbers}-{20m+5}-{20m+15} because for these p LHSmod(25) \neq0. So the set of p's for which n=1 is true is strictly larger than the prime set.


    I really have no idea what you mean by this and why you think this proves anything even close to the OP. In fact, I also

    have no idea why would anyone think that what you sent in your past messages, including the table, is a proof of anything.

    Tonio



    Actually, I suspect that n=1 is true for ANY positive integer p! However a proof for that might be much more involved and I shall not attempt to prove it
    .
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    Re: Equation with primes in exponent

    Quote Originally Posted by tonio View Post
    I really have no idea what you mean by this and why you think this proves anything even close to the OP. In fact, I also

    have no idea why would anyone think that what you sent in your past messages, including the table, is a proof of anything.

    Tonio
    Of course this proves the question!! and more than the question!

    Let me try to clarify my reasoning a little for you:
    1) {2}^{p}+{3}^{p}=0mod(5) for p=odd. We've shown this, right?

    2) Now RHS ={a}^{n}. If RHS=LHS, then RHS=0mod(5). If n>2, we must also have RHS=0mod(25). So LHS=0mod(25). So if we can show LHS \neq0mod(25) for prime p's, then we are done (this is also what abhishekkgp claimed up there).

    3) But x=0mod(25) iff last two digits of x are 00, 25, 50 or 75. But LHS ={2}^{p}+{3}^{p} has those last two digits iff p=20m+5 or p=20m+15, which are NOT primes! Hence prime for prime p LHS \neq0(mod25). and we're done!
    -------------------------------------------------------

    Why the table is enough? Because there's only FINITE p's we need to check (because we only care about last two digits!)

    EDIT: Red letters
    Last edited by godelproof; June 23rd 2011 at 04:33 AM.
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    Re: Equation with primes in exponent

    Quote Originally Posted by godelproof View Post
    Of course this proves the question!! and more than the question!

    Let me try to clarify my reasoning a little for you:
    1) {2}^{p}+{3}^{p}=0mod(5) for p=odd. We've shown this, right?

    2) Now RHS ={a}^{n}. If RHS=LHS, then RHS=0mod(5). If n>2, we must also have RHS=0mod(25). So LHS=0mod(25). So if we can show LHS \neq0mod(25) for prime p's, then we are done (this is also what abhishekkgp claimed up there).

    3) But x=0mod(25) iff last two digits of x are 00, 25, 50 or 75. But LHS ={2}^{p}+{3}^{p} has those last two digits iff p=20m+5 or p=20m+15,



    This is my point: where do you think you proved this?? I'm not contending the other rather obvious facts.



    which are NOT primes! Hence prime for prime p LHS \neq0(mod25). and we're done!
    -------------------------------------------------------

    Why the table is enough? Because there's only FINITE p's we need to check (less than 100 because we only care about last two digits!)


    Excuse me? What does "to care about the last two digits" have to do with only "having" to check a finite number of primes? Is this something you

    already proved?

    In fact, in your "proof" (I'll take the marks off when I'm convinced it really is one) it really is even

    easier: the last two digits of RHS have to be either 25 or 75. You can dispose of 00 and 50, but then

    one has to prove (and I think this was your intention) that it is enough to prove the claim for

    a finite number of primes because so and so. I just can't see it.

    Tonio


    .
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    Re: Equation with primes in exponent

    Quote Originally Posted by tonio View Post
    In fact, in your "proof" (I'll take the marks off when I'm convinced it really is one) it really is even

    easier: the last two digits of RHS have to be either 25 or 75. You can dispose of 00 and 50, but then

    one has to prove (and I think this was your intention) that it is enough to prove the claim for

    a finite number of primes because so and so. I just can't see it.

    Tonio
    Define LT(x)=last two digits of x (so LT(100)=00, LT(234)=34, ect.). Then LT(2^5+3^5)=75. Define {k}^{*} to be the smallest integer>5 such that 1) LT(2^ {k}^{*} +3^ {k}^{*} )=00 or 25 or 50 or 75; 2) LT(2^ {k}^{*} )=LT(2^5); 3) LT(3^ {k}^{*} )=LT(3^5). Such a {k}^{*} is found by inspecting the first 20 terms of the table to be {k}^{*} =25. And LT(2^ {k}^{*} +3^ {k}^{*} )=75. No problem with this?

    Now because LT function takes only the last two digits, it is true that LT(2^(5+i))=LT ({2}^{5}{2}^{i})=LT(LT(2^5)LT(2^i))=LT(LT(2^ {k}^{*})LT(2^i))=LT( {2}^{{k}^{*}}{2}^{i})=LT( 2^{{k}^{*}+i}), for any i>0. Similarly LT(3^(5+i))=LT( 3^{{k}^{*}+i}), too. So LT(2^(5+i)+3^(5+i))=LT(LT(2^(5+i))+LT(3^(5+i)))=LT (LT(2^( {k}^{*}+i))+LT(3^( {k}^{*}+i)))=LT(2^( {k}^{*}+i)+3^( {k}^{*}+i))

    From this we can see f(k)=LT(2^k+3^k) takes only FINITELY MANY values for k=5,6,7,...
    and the values it can take are exactly equal to this set:{f(5),f(6),f(7),...,f( {k}^{*}-1)}, and any f(k) \in{00,25,50,75} iff there exists 5 \leqk' \leq {k}^{*}-1 such that f(k')=f(k).

    So what is this k' if f(k') \in{00,25,50,75}? Inspect the table we find k'=5 or k'=15.

    So now you see where k=5+20m and k=15+20m come from.
    --------------------------------------------------------------
    If this doesn't convince you. I happily quit

    ---------------------------------------------------------
    EDIT: red letters.
    Last edited by godelproof; June 23rd 2011 at 06:04 AM.
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